### Video Transcript

Consider the following graph of π¦
equals negative three over π₯ minus two. By looking at the graph and
substituting a few successively larger values of π₯ into the function, what is the
end behavior of the graph as π₯ increases along the positive π₯-axis?

In this question, weβre given the
graph of a rational function π¦ equals negative three over π₯ minus two and asked to
use this graph and the function to determine the end behavior of its graph. To do this, we can first recall
that the end behavior of a graph is the behavior of the graph as the values of π₯
increase or decrease without bound. We want to determine the end
behavior of the graph as the values of π₯ increase without bounds. And we are told to analyze the end
behavior using two different methods. First, we are told to use the given
graph to analyze the end behavior. Second, we are told to analyze the
end behavior by substituting successively larger values of π₯ into the given
function.

Letβs start by looking at the given
graph. We recall that the π₯-coordinate of
a point on the graph represents the input value of the function and the
π¦-coordinate represents the corresponding output of the function. From the graph, we can see that as
our values of π₯ get larger and larger, the graph of the function approaches the
π¦-axis from below. This appears to be a horizontal
asymptote at π¦ equals zero. Since the graph approaches the line
π¦ equals zero as the values of π₯ increase without bound, this implies that the
value of π¦ approaches zero as π₯ increases.

This is enough to answer the
question. However, we are asked to check this
answer by substituting successively larger input values into the function. To do this, letβs first look at the
rational function. We note that the denominator is π₯
minus two. This means that we can make
calculations easier by making the denominator easier to evaluate. For example, if we substitute π₯
equals 12 into the function, then the denominator will evaluate to give 10, and
division by 10 is easy to calculate. We have that when π₯ equals 12, π¦
is equal to negative three divided by 12 minus two, which we can evaluate to give
negative 0.3.

We can follow the same process with
a larger input value. This time we will use π₯ equals
102. This time, the denominator
evaluates to give 100, and we can calculate that when π₯ is equal to 102, π¦ is
equal to negative 0.03. This pattern continues
indefinitely. For example, we can calculate that
when π₯ is 1,002, π¦ is equal to negative 0.003. We see that as we are increasing
the input values of π₯, the magnitude of the output values of π¦ are getting smaller
and smaller. In fact, they are approaching the
value zero from the negative direction. This agrees with our analysis from
the graph. So we can conclude that as the
values of π₯ increase along the positive π₯-axis without bound, the values of π¦
will approach zero.